Second Bianchi identity

Statement

Let $\nabla$ be a linear connection and $R$ be the Riemann curvature tensor of $\nabla$. Then $R$ can itself be differentiated via $\nabla$, since $R$ is a $(1,3)$-tensor and we can define the connection on all $(p,q)$-tensors. With this meaning, the following cyclic summation is zero:

$(\nabla _{X}R)(Y,Z)+(\nabla _{Y}R)(Z,X)+(\nabla _{Z}R)(X,Y)=0$

Proof

To prove this we look more closely at what $\nabla _{X}R$ means.

$\nabla _{X}R$ must satisfy the following compatibility condition:

$((\nabla _{X}R)(Y,Z))(W)+R(\nabla _{X}Y,Z)W+R(Y,\nabla _{X}Z)W+R(Y,Z)\nabla _{X}W=\nabla _{X}(R(Y,Z)W)$

We now concentrate on all the other terms.